Thread: Help with nonlinear equation - reduction of order

Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
(y^3)y''=1

So far, using a substitution of p=y', I have managed to do:
(y^3)p(dp/dy)=1
p dp = 1/(y^3) dy
(p^2)/2 = -1/(2y^2) + c
p^2 = -1/(y^2) + c

Can someone please confirm if my working so far is correct or incorrect and where to go from here.

Thanks

Originally Posted by dwally89

Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
(y^3)y''=1

So far, using a substitution of p=y', I have managed to do:
(y^3)p(dp/dy)=1
p dp = 1/(y^3) dy
(p^2)/2 = -1/(2y^2) + c
p^2 = -1/(y^2) + c

Can someone please confirm if my working so far is correct or incorrect and where to go from here.

Thanks

So far so good...


$\displaystyle p=\sqrt{c -\frac{1}{y^2} }$

$\displaystyle p=\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }$

$\displaystyle dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}} $

Now take integral on both sides.

Originally Posted by Also sprach Zarathustra

So far so good...


$\displaystyle p=\sqrt{c -\frac{1}{y^2} }$

$\displaystyle p=\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }$

$\displaystyle dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}} $

Now take integral on both sides.

I managed to get to that step, but am not sure how to take the integral of the right hand side.

Thanks

Originally Posted by dwally89

I managed to get to that step, but am not sure how to take the integral of the right hand side.

Thanks

$\displaystyle \int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy$

Now, substitute $\displaystyle u=cy^2-1$ then $\displaystyle du=2cy$

Try to finish...

Originally Posted by Also sprach Zarathustra

$\displaystyle \int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy$

Now, substitute $\displaystyle u=cy^2-1$ then $\displaystyle du=2cy$

Try to finish...

So:
$\displaystyle y=\sqrt{cx^2+\frac{1}{c}}$

which then working backwards, to check that it works gives:
$\displaystyle y^3=({cx^2+\frac{1}{c}})^{1.5}$
$\displaystyle y''=({cx^2+\frac{1}{c}})^{-1.5}$
$\displaystyle y^3y''=1$

Right?

Originally Posted by dwally89

So:
$\displaystyle y=\sqrt{cx^2+\frac{1}{c}}$

which then working backwards, to check that it works gives:
$\displaystyle y^3=({cx^2+\frac{1}{c}})^{1.5}$
$\displaystyle y''=({cx^2+\frac{1}{c}})^{-1.5}$
$\displaystyle y^3y''=1$

Right?

I don't think so...



Steps after taking integration on both sides:

$\displaystyle \frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B$

Or:

$\displaystyle y\sqrt{c-\frac{1}{y^2}}=x+K$


Or:

$\displaystyle \sqrt{cy^2-1}=x+K$

Or:


$\displaystyle cy^2-1=(x+K)^2$


Now solve for $\displaystyle y$.

Originally Posted by Also sprach Zarathustra

I don't think so...



Steps after taking integration on both sides:

$\displaystyle \frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B$

Or:

$\displaystyle y\sqrt{c-\frac{1}{y^2}}=x+K$


Or:

$\displaystyle \sqrt{cy^2-1}=x+K$

Or:


$\displaystyle cy^2-1=(x+K)^2$


Now solve for $\displaystyle y$.

I have a bad habit of forgetting to add constants when integrating...

$\displaystyle y=\sqrt{\frac{(x+K)^2+1}{c}}$
Is that right?

Originally Posted by dwally89

$\displaystyle y=\sqrt{\frac{(x+K)^2+1}{c}}$
Is that right?

$\displaystyle y=\sqrt{\frac{(x+K)^2+1}{c}}$ or $\displaystyle y=-\sqrt{\frac{(x+K)^2+1}{c}}$

Thanks